For the carbon, we start with 4 valence electrons, we have 0 lone pair electrons minus 4, and we end up with a formal charge of 0.
对于碳,我们从四个价电子开始,我们有零个孤对电子,再减去四,最终我们有零个形式电荷。
And he gets, you are absolutely right, 00163 it is not 4.0, it is 4.00163. Take that.
他得到的结果是,你完全是对的,不是4。0,是4。
So if I have 32 bits, each of which can be a 0 or 1, that's two possibilities for every place, so 2 to the 32, that's 4 billion.
如果有32比特的话,每个比特都有0或,两种可能性,所以是2的32次方,也就是40亿。
Middle does best against Right, after all, 4 is bigger than 0 and 4 is bigger than 2, so Middle does best against Right.
选中是右的最佳对策是吧,因为,4大于0,4大于2,所以选中是右的最佳对策
So for the bond order we want to take 1/2 of the total number of bonding electrons, so that's going to be 4 minus anti-bonding is 4, so we end up getting a bond order that's equal to 0.
键序等于1/2乘以,总的成键电子数,也就是4,减去反键电子数,也就是4。,所以最后得到键序为0。
to .4, so .2 is here in the middle; that's .2.
到0.4,那么0.2在中点,这就是0.2
Again I should have said first, index 0, the first one. I can similarly go in and say I'd like all the things between index 2 and index 4. And again, remember what a b c that does. Index 2 says start a 0. 1, 2. So a, b, c.
我还是要说一遍,索引为0的元素,是第一个元素,我可以要求返回索引,2和4之前的所有元素,请记住2,是从0开始的,那么0,1,2对应的是。
Let's do this from 0 to .4.
这里是0到0.4
No, instead we have to click other amount at bottom right there, when we have to input 4-0-0 in this screen 4-0-0 after consulting visually the little cheat sheet on the device itself it tells you in a very long chart how much one pass costs so we multiply and type in 4-0-0, we hit enter.
不,然而我们必须点击,右下边的这个,当我们需要在这个屏幕上,输入0,通过咨询那个设备上小说明,它在一个很长的图表上,告诉你,每一站需要多少钱,我们加起来,输入了4-0-0,点击输入。
, okay. So 1 plus 1/4 of 0 is 1, so if Player II chooses 0, player I's best response is to choose 1.
是1,因为1+0/4=1,参与人II选0时I的最佳对策是1
Then, if n = 100--now I'm going to label this x differently, I'm now going to show the normal bell-shaped curve and I'm going to do this from 0 to .4.
如果n=100,现在我得重新标x值,我现在要画正态分布的钟形曲线了,在这儿用0到0.4
Now I'll print L 2. Kind of what you'd guess, but here's the interesting question: if I say L 1 is assigned 0, L 1 sub is assigned 4, I'll print L 1.
然后我将显示下L2的内容,你,们猜猜这个有趣的问题:,如果我将L1赋值为4的话,再去显示下L1,会怎么样呢?
So with an integer, an int data type, you can store any number between 0 and 4 billion roughly.
只要有一个整数,一个int型的数据,就能储存任意一个,位于0到40亿之间的一个数。
so in some sense it's safer, it avoids the 0, although I mean it's true it avoids the 0 but it also doesn't get the 5 and the 4.
某些情况下选下是比较安全的,因为他避免了0收益,我的意思是它的确避免了0收益,但选它也得不到5和4这连个最大收益
You could choose any real number in the interval .
你可以选择0到4间的任意实数
So the strategy choices, we're not going to do it in hours, let's just normalize and regard these choices as living in 0 to 4, and you can choose any number of hours between 0 and 4.
我们用小时数来表示策略吧,用0到4的数字来标记这些策略,你可以从0到4中任选工作的小时数
So we said B was somewhere between 0 and 1/4, let's draw the case for B equals 1/4.
我们之前也说过B在0到1/4之间,这里我们绘制B=1/4的情况吧
We'll assume that B lives between 0 and a 1/4 and it's known, I just want to be able to vary it later.
假设B是0到1/4之间的已知量,我希望一会能改变一下这个数据
You can search obviously by any of the fields up top by days ; of the week you want a class; days of the week you don't want a class; grad level or undergrad level, faculty scores that are 4.0 and higher, ; course overall scores that are 4.0 and higher; and a whole number of other features including synchronization with Google calendar.
你可以按日期来搜索网页上的任一科目,以决定哪天去上课,哪天不去上课;,研究生课程或本科生课程,教师打分为4分或更高的课程,课程总学分为4分或者更高的课程;,当然很多其他选择方式,包括Google同步日历。
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